Faraday's law — EMF = −dΦ - dt: Worked examples

1.8.26 · D3 · Physics › Electromagnetism › Faraday's law — EMF = −dΦ - dt

Definition Woh chaar symbols jo hum use karte rahenge

$B$ — magnetic field strength, tesla (T) mein maapa jaata hai. Isse socho jaise andekhey dhaagon ki density jo space mein guzarti hain, jaise har square metre mein spaghetti ke dhaagon ki ginatri.
$A$ — flat loop ki area, square metres ( $m^{2}$ ) mein. Woh hoop kitna bada hai jisme dhaage guzarte hain.
$θ$ — field $B$ aur loop ke normal $\overset{n}{^}$ ke beech ka angle (ek arrow jo loop ke face se seedha bahar nikaala hua hai, uske perpendicular). Jab loop field ki taraf seedha face karta hai, $θ = 0$ .
$N$ — wire ke turns ki ginatri, yani wahi wire kitni baar loop karta hai pehle ki dhaagon ko phir se count kiya jaaye. 50 baar winded coil mein $N = 50$ hoga, aur wahi flux 50 baar "feel" hoga. Ek single loop mein $N = 1$ hota hai.

Milaakar, flux hai $Φ_{B} = B A cos θ$ — "kitne dhaage actually hoop mein se guzar rahe hain." Iska unit hai weber, jiska symbol $Wb$ hai; ek weber barabar hai $1 T \cdot m^{2}$ .

Scenario matrix

Faraday's law hai $E = - \frac{d Φ _{B}}{d t}$ , aur $Φ_{B} = B A cos θ$ mein teen knobs hain ( $B$ , $A$ , $θ$ ). Har problem ek ya zyada knobs ka ghoomna hai, plus ek sign question. Yeh poora grid hai. Das cells; Example 4 ek saath do cells cover karta hai (dono degenerate zeros), isliye naau examples saare das fill karte hain:

Cell	Case class	Kya badalta hai	Special feature	Example
C1	Changing field	$B (t)$	linear ramp, area fixed	Ex 1
C2	Changing area	$A (t)$	sliding rod (motional)	Ex 2
C3	Rotating loop	$θ (t)$	sine output, sign per quadrant	Ex 3
C4	Degenerate: edge-on	$θ = 90°$	flux zero hai	Ex 4
C5	Degenerate: constant	kuch nahi badalta	EMF zero hai	Ex 4
C6	Sign / direction	Lenz	current kis taraf flow karta hai	Ex 5
C7	Two knobs at once	$B (t)$ aur $A (t)$	product rule, terms add hote hain	Ex 6
C8	Non-linear in time	$B \propto t^{2}$	EMF khud time ke saath badalta hai	Ex 7
C9	Real-world word problem	card / coil swipe	words ko knobs mein translate karo	Ex 8
C10	Exam twist: limiting value	field switched off in $Δ t \to 0$	spike / average EMF	Ex 9

Example 1 — Changing field, fixed loop `(cell C1)`

Pehchano kaunsa knob ghoom raha hai. Loop move nahi karta ( $A$ fixed) aur tilt nahi hota ( $θ = 0$ fixed). Sirf $B$ badal raha hai. Yeh step kyun? Spaghetti dhaagon ko imagine karo jo denser hote ja rahe hain jabki hoop wahi khada hai — zyada dhaage usi opening se thunste ja rahe hain. Ek moving knob ko naam dene se product-rule ke do terms khatam ho jaate hain, sirf $\frac{d B}{d t} A cos θ$ bachta hai.
Area nikalo. $A = s^{2} = (0.20)^{2} = 0.040 m^{2}$ . Yeh step kyun? Area chahiye taaki field-rate ko flux-rate mein badla ja sake — loop ka size decide karta hai ki $B$ ki har unit rise kitne extra dhaage push karegi.
Field ka rate of change nikalo. Steady ramp matlab $\frac{d B}{d t} = \frac{0.80 - 0}{4.0} = 0.20 T/s$ . Yeh step kyun? "Steadily" matlab constant slope, toh rise-over-run exact rate deta hai; seedhi line ke liye koi calculus nahi chahiye.
Faraday's law apply karo (magnitude, $cos 0 = 1$ ): $∣ E ∣ = N A \frac{d B}{d t} = 50 \times 0.040 \times 0.20 = 0.40 V .$ Yeh step kyun? $N = 50$ turns mein se har ek same flux link karta hai, toh har turn apna chota EMF add karta hai aur woh series mein stack ho jaate hain.

Example 2 — Sliding rod, changing area `(cell C2)`

Swept area dekho (Figure 1 mein mint strip). Jab rod right ki taraf $d x$ move karta hai, loop ka area $d A = L d x$ se badhta hai. Yeh step kyun? Field ko page par khade vertical spaghetti dhaagon ka carpet socho. Jab rod rightward sweep karta hai toh woh naye dhaage rake in karta hai; mint area exactly woh patch hai jo ab loop ke andar pakda gaya hai. $B$ aur $θ$ kabhi nahi badte — sirf area badlta hai — isliye yeh Motional EMF hai.
Rate of area change. $\frac{d A}{d t} = L \frac{d x}{d t} = Lv = 0.25 \times 4.0 = 1.0 m^{2} / s$ . Yeh step kyun? Speed exactly hai "kitni tez $x$ badh raha hai," toh $d x / d t = v$ ; rod ki length se multiply karo area swept per second paane ke liye.
Faraday deta hai $∣ E ∣ = B \frac{d A}{d t} = B Lv = 0.30 \times 0.25 \times 4.0 = 0.30 V .$ Yeh step kyun? Field density $B$ times new-area-rate = new-strand-rate = flux rate. Parent ke $B v L$ force-on-charges derivation jaisi hi number — do pictures, ek answer.
Current $I = \frac{E}{R} = \frac{0.30}{1.5} = 0.20 A .$ Yeh step kyun? EMF $0.30 V$ ki battery ki tarah kaam karta hai jo resistor mein current push karta hai (Ohm's law).

Example 3 — Rotating coil (generator), sign per quadrant ke saath `(cell C3)`

Time reference set karo, phir flux likho. Humne chose kiya $t = 0$ par $θ = 0$ , toh $θ = ω t$ aur $Φ_{B} = B A cos (ω t)$ . Yeh step kyun? Phase depend karta hai ki clock kahaan start kiya. Agar loop edge-on se shuru karte ( $t = 0$ par $θ = 90°$ ), toh likhte $Φ_{B} = B A cos (ω t + 90°) = - B A sin (ω t)$ , aur EMF $cos$ ke roop mein aata — shape ek jaisi hai, sirf shifted, aur alag starting phase flip kar deta hai $t = 0$ par EMF ki sign. Reference batana hi sign ko pin karta hai.
Differentiate karo. $E = - N \frac{d}{d t} (B A cos ω t) = N B A ω sin ω t$ . Yeh step kyun? $cos$ ka derivative $- sin$ hai; do minus signs (Faraday ka aur derivative ka) cancel ho jaate hain, clean $+ sin$ bachta hai. Yeh sine isliye hai ki generators AC dete hain — dekho Electric Generators and AC.
Peak value (jab $sin = 1$ ): $E_{0} = N B A ω = 200 \times 0.40 \times 0.015 \times 100 = 120 V .$ Yeh step kyun? $sin$ maximum $1$ hota hai; peak wave ka amplitude hai.
$θ = 60°$ par: $E = E_{0} sin 60° = 120 \times 0.8660 = 103.9 V .$ Yeh step kyun? Instantaneous EMF amplitude times $sin θ$ hai.
Har quadrant mein sign. $θ$ ke $0 \to 360°$ ke saath:
- $θ \in (0°, 180°)$ : $sin θ > 0$ , toh $E > 0$ — flux gir raha hai, current ek taraf.
- $θ \in (180°, 360°)$ : $sin θ < 0$ , toh $E < 0$ — flux badh raha hai, current reverse.
Yeh step kyun? Yahi AC ka sign trap hai: same magnitude har turn mein do baar aata hai lekin opposite current direction se (Lenz track karta hai). $θ = 240°$ par, misal ke taur par, $sin 240° = - 0.866$ , toh $E = - 103.9 V$ — $60°$ jaisa hi size, opposite sign.

Example 4 — Do degenerate cases `(cells C4 & C5)`

Case (a) — constant field. $\frac{d B}{d t} = 0$ , area fixed, angle fixed. Product rule ka har term zero hai. $E = 0 V .$ Yeh step kyun? EMF change track karta hai, flux ki size nahi. Ek bahut bada frozen field kuch induce nahi karta — yahi parent ki headline mistake hai ("big flux ⇒ big EMF" galat hai).
Case (b) — edge-on loop. Yahan $θ = 90°$ , toh $cos θ = 0$ . Flux hai $Φ_{B} = B A cos 90° = 0$ har instant par, chahe $B$ kaise bhi badhe. $Φ_{B} (t) = 0 \Rightarrow \frac{d Φ _{B}}{d t} = 0 \Rightarrow E = 0 V .$ Yeh step kyun? Field loop ke plane ke saath sliding karta hai toh kuch bhi hoop mein se nahi guzarta, toh chahe $B$ kitna bhi tez badhe, zero dhaage hoop se guzarte hain. $cos θ$ factor exactly wahi kaam kar raha hai jo parent ne describe kiya tha.

Example 5 — Current ki sign / direction `(cell C6)`

Magnitude. $∣ E ∣ = \frac{d Φ _{B}}{d t} = 0.60 V .$ Yeh step kyun? Ek turn, single knob — seedha application.
Lenz se direction (dekho Lenz's Law). Downward flux badh raha hai. Loop change oppose karta hai, toh loop ke andar apna magnetic field upar pointing banata hai, increase resist karne ke liye. Yeh step kyun? Faraday's law mein minus sign yahi opposition hai — energy conservation forbid karta hai loop ko change help karne se.
Current ke liye right-hand rule. Right hand curl karo taaki thumb upar point kare (loop ka induced field); fingers counterclockwise curl karti hain upar se dekha hua. Yeh step kyun? Current direction aur uska magnetic field right-hand rule se lock hain saath mein.

Example 6 — Ek saath do knobs ghoomna `(cell C7)`

Product rule. $θ = 0$ ke saath (toh $cos θ = 1$ aur uska rate zero hai), sirf do terms bachte hain: $\frac{d Φ _{B}}{d t} = \frac{d B}{d t} A + B \frac{d A}{d t} .$ Yeh step kyun? Do knobs ghoom rahe hain matlab do product-rule terms; hum single-term shortcut use nahi kar sakte.
Plug in karo. $\frac{d Φ _{B}}{d t} = (0.40) (0.050) + (0.30) (0.010) = 0.020 + 0.003 = 0.023 Wb/s .$ Yeh step kyun? Har term ko given instant par evaluate karo, phir add karo — dono effects flux badhate hain, toh reinforce karte hain.
EMF: $∣ E ∣ = 0.023 V = 23 mV .$

Example 7 — Non-linear field in time `(cell C8)`

Derivative kyun, slope kyun nahi. $B$ ab ek curve hai, seedhi ramp nahi, toh rate $\frac{d B}{d t}$ har instant par alag hoga — differentiate karna zaroori hai. Yeh step kyun? "Rise over run" sirf seedhi lines ke liye kaam karta hai; yahan slope change hota rehta hai.
Differentiate karo. $\frac{d B}{d t} = \frac{d}{d t} (k t^{2}) = 2 k t .$ Yeh step kyun? Power rule: $t^{2}$ ka derivative $2 t$ hai. Isliye EMF constant nahi hai — yeh $t$ ke saath linearly badh raha hai.
EMF magnitude. $∣ E ∣ = A \frac{d B}{d t} = A \cdot 2 k t .$ $t = 3.0 s$ par: $∣ E ∣ = 0.020 \times 2 \times 0.50 \times 3.0 = 0.060 V .$

Example 8 — Real-world word problem `(cell C9)`

Words ko ek knob mein translate karo. Coil tilt ya resize nahi karti; field jo woh dekhti hai drop ho jaati hai. Toh yeh $\frac{d B}{d t}$ (changing-field) case hai, cell C1 ka cousin. Yeh step kyun? Word problems sirf disguise mein ek knob hain; naam lene se sahi formula mil jaata hai. Yahan dhaage jo coil dekhti hai suddenly gayab ho jaate hain jab woh strip clear karti hai.
Average field-rate. $\frac{Δ B}{Δ t} = \frac{2.0 \times 1 0 ^{- 3}}{5.0 \times 1 0 ^{- 4}} = 4.0 T/s .$ Yeh step kyun? "Average" matlab total change ko total time par use karo, instantaneous derivative nahi — field smooth formula ke roop mein nahi diya gaya.
Faraday's law mein plug karo. $∣ \overset{ˉ}{E} ∣ = N A \frac{Δ B}{Δ t} = 300 \times (1.0 \times 1 0^{- 4}) \times 4.0 = 0.12 V .$ Yeh step kyun? Same $N A d B / d t$ machinery Ex 1 jaisi — coil fixed hai, sirf field change hoti hai, aur $N = 300$ turns EMF stack karte hain taaki tiny per-turn signal readable ban jaye.
Magnitude sense check karo. $0.12 V = 120 mV$ — hundreds of millivolts, na microvolts na volts. Yeh step kyun? Quick order-of-magnitude check slip hue powers of ten pakad leta hai; $120 mV$ ek aisa signal hai jo real reader detect kar sakta hai.

Example 9 — Limiting spike `(cell C10)`

Total flux change. $∣Δ Φ_{B} ∣ = A ∣Δ B ∣ = 0.10 \times 0.60 = 0.060 Wb .$ Yeh step kyun? Flux $B A$ se $0$ tak drop hota hai; change ki maatraa fixed hai chahe kitni bhi tezi se karo — wahi dhaagon ka bundle hoop se jaana hai.
(a) ke liye average EMF. $∣ \overset{ˉ}{E} ∣ = \frac{∣Δ Φ _{B} ∣}{Δ t} = \frac{0.060}{0.20} = 0.30 V .$ Yeh step kyun? Average EMF = total flux change spread over switch-off time.
(b) ke liye limit. Numerator $∣Δ Φ_{B} ∣ = 0.060 Wb$ fixed hai, toh $∣ \overset{ˉ}{E} ∣ = \frac{0.060}{Δ t} Δ t \to 0 \infty.$ Yeh step kyun? Ek fixed number ko hamesha chhote hote time se divide karna blow up karta hai. Real circuits mein coil ki self-inductance is sudden change ko resist karti hai aur tumhe ek bada voltage spike milta hai — isliye coil par switch open karne se spark aa sakta hai.

Coverage check — kya humne har cell hit kiya?

Recall Har matrix cell, apne example ke saath

C1 changing field ::: Ex 1 (aur Ex 8 ka cousin) C2 changing area / motional ::: Ex 2 C3 rotating loop / generator + sign per quadrant ::: Ex 3 C4 degenerate edge-on ( $θ = 90°$ ) ::: Ex 4(b) C5 degenerate constant field ::: Ex 4(a) C6 sign / current direction (Lenz) ::: Ex 5 C7 two knobs at once (product rule) ::: Ex 6 C8 non-linear in time ( $B \propto t^{2}$ ) ::: Ex 7 C9 real-world word problem ::: Ex 8 C10 limiting value / spike ::: Ex 9

Connections

Magnetic Flux — woh quantity $Φ_{B} = B A cos θ$ jo hum throughout differentiate karte hain
Motional EMF — Ex 2 ke sliding rod ke peeche ki physics
Lenz's Law — direction rule jo Ex 5 mein use hoti hai
Electric Generators and AC — Ex 3 ka sine output
Inductance — kyun Ex 9 ka spike real coils mein finite hota hai
Maxwell's Equations — Faraday's law field form mein
Hinglish parent note

Worked examples — Faraday's law — EMF = −dΦ - dt